OCR Further Statistics 2018 December — Question 4 8 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2018
SessionDecember
Marks8
TopicPoisson distribution
TypeExplain or apply conditions in context
DifficultyModerate -0.3 This is a straightforward application of Poisson distribution with standard bookwork parts: (a) is pure recall of conditions, (b) is routine probability calculation with rate adjustment, (c) applies binomial expectation, and (d) requires basic comparison of observed vs expected. The multi-part structure and rate scaling add slight complexity, but all techniques are standard Further Stats exercises requiring no novel insight.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities

4 Leyla investigates the number of shoppers who visit a shop between 10.30 am and 11 am on Saturday mornings. She makes the following assumptions.
  • Shoppers visit the shop independently of one another.
  • The average rate at which shoppers visit the shop between these times is constant.
    1. State an appropriate distribution with which Leyla could model the number of shoppers who visit the shop between these times.
Leyla uses this distribution, with mean 14, as her model.
  • Calculate the probability that, between 10.35 am and 10.50 am on a randomly chosen Saturday, at least 10 shoppers visit the shop. Leyla chooses 25 Saturdays at random.
  • Find the expected number of Saturdays, out of 25, on which there are no visitors to the shop between 10.35 am and 10.50 am .
  • In fact on 5 of these Saturdays there were no visitors to the shop between 10.35 am and 10.50 am . Use this fact to comment briefly on the validity of the model that Leyla has used.

  • (a)
    Poisson
    AnswerMarks
    B1(no additional guidance)
    (b)
    \(\lambda = 7\)
    \(P(\geq 10) = 1 - Po(\leq 9) = 1 - 0.8305\)
    \(= 0.1695\)
    AnswerMarks
    B1, M1, A1Allow for \(1 - 0.9015 = 0.0985\); Not Poisson: allow B1M1
    (c)
    \(P(0) = 0.000912\)
    \(E(\text{number}) = 25 \times P(0)\)
    \(= 0.0228\)
    AnswerMarks
    B1, M1, A1In range 0.0225, 0.0235]; Not Poisson: allow B1M1
    (d)
    Suggests that the model is not valid
    AnswerMarks
    B1Allow from other distributions
    ## (a)
    Poisson
    
    | B1 | (no additional guidance) |
    
    ## (b)
    $\lambda = 7$
    
    $P(\geq 10) = 1 - Po(\leq 9) = 1 - 0.8305$
    
    $= 0.1695$
    
    | B1, M1, A1 | Allow for $1 - 0.9015 = 0.0985$; Not Poisson: allow B1M1 |
    
    ## (c)
    $P(0) = 0.000912$
    
    $E(\text{number}) = 25 \times P(0)$
    
    $= 0.0228$
    
    | B1, M1, A1 | In range 0.0225, 0.0235]; Not Poisson: allow B1M1 |
    
    ## (d)
    Suggests that the model is not valid
    
    | B1 | Allow from other distributions |
    
    ---
    4 Leyla investigates the number of shoppers who visit a shop between 10.30 am and 11 am on Saturday mornings. She makes the following assumptions.
    
    \begin{itemize}
      \item Shoppers visit the shop independently of one another.
      \item The average rate at which shoppers visit the shop between these times is constant.
    \begin{enumerate}[label=(\alph*)]
    \item State an appropriate distribution with which Leyla could model the number of shoppers who visit the shop between these times.
    \end{itemize}
    
    Leyla uses this distribution, with mean 14, as her model.
    \item Calculate the probability that, between 10.35 am and 10.50 am on a randomly chosen Saturday, at least 10 shoppers visit the shop.
    
    Leyla chooses 25 Saturdays at random.
    \item Find the expected number of Saturdays, out of 25, on which there are no visitors to the shop between 10.35 am and 10.50 am .
    \item In fact on 5 of these Saturdays there were no visitors to the shop between 10.35 am and 10.50 am . Use this fact to comment briefly on the validity of the model that Leyla has used.
    \end{enumerate}
    
    \hfill \mbox{\textit{OCR Further Statistics 2018 Q4 [8]}}